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Math Help - Help needed please!

  1. #1
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    Help needed please!

    A rectangle made of elastic material will be made into a cylinder by joining edges AD and BC. To support the structure, a wire of fixed length L is placed along the diagonal of the rectangle. Find the angle \theta that will result in the cylinder of maximum volume.

    see attached picture

    thank you very much!

    unfortunately i cannot attach a picture
    so here is a shot at it...

    d._________c
    ..|............/|
    ..|........./...|
    ..|....../......|
    ..|.../.........|
    a|/............|b
    theta is angle a, in the triangle cab
    Last edited by doctorgk; April 26th 2008 at 03:03 PM.
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  2. #2
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    Hello, doctorgk!

    A rectangle made of elastic material will be made into a cylinder by joining edges AD and BC.
    To support the structure, a wire of fixed length L is placed along the diagonal of the rectangle.
    Find the angle \theta\;(\angle CAB) that will maximize the volume of the cylinder.
    Code:
        D * - - - - - - - * C
          |             * |
          |           *   |
          |      L  *     |
          |       *       | L·sinθ
          |     *         |
          |   *           |
          | * θ           |
        A * - - - - - - - * B
               L·cosθ
    We have: . AB \,=\,L\cos\theta,\;CB \,=\,L\sin\theta

    The height of the cylinder is: . h \:=\:CB \:=\:L\sin\theta\;\;{\color{blue}[1]}

    AB is the circumference of the circular base of the cylinder.
    . . L\cos\theta \,=\,2\pi r\quad\Rightarrow\quad r \:=\:\frac{L\cos\theta}{2\pi}\;\;{\color{blue}[2]}


    The volume of a cylinder is: . V \;=\;\pi r^2h\;\;{\color{blue}[3]}

    Substitute [1] and [2] into [3]: . V \;=\;\pi\left(\frac{L\cos\theta}{2\pi}\right)^2(L\  sin\theta)

    . . and we have: . \boxed{V \;=\;\frac{L^3}{4\pi}\,\sin\theta\cos^2\!\theta}


    And that is the function to be maximized . . . good luck!

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  3. #3
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    So far

    i got it so far, but just a quick question on the optimization. when u differentiate would you factor out the (L^3)/4 pi and then differentiate the trig functions, or would you differentiate the entire thing?
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  4. #4
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    Hello,

    I think it's better factorising, less obstruction, more visibility
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  5. #5
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    what do i do next?

    so i got the derivative to be...
    \frac{dV}{d\theta}=(\frac{L^3}{4\pi})\cos^3\theta - \sin\theta2\cos\theta\sin\theta

    then i set it equal to zero and i get

    0=\cos^3\theta - \sin\theta2\cos\theta\sin\theta

    what do i do from here?
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  6. #6
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    Quote Originally Posted by doctorgk View Post
    so i got the derivative to be...
    \frac{dV}{d\theta}=(\frac{L^3}{4\pi}) {\color{red}(}\cos^3\theta + \sin\theta2\cos\theta\-sin\theta{\color{red})}

    then i set it equal to zero and i get

    0=\cos^3\theta + \sin\theta2\cos\theta

    what do i do from here?
    Hm... No... You have to keep the factor L^3/4pi for the whole thing (correction in red)

    The derivative will be :
    \underbrace{\cos^3(\theta)}_{\text{You found it}}+\sin(\theta) 2 (\underbrace{-\sin(\theta)}_{\text{derivative of cos}}) \cos(\theta)
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  7. #7
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    then i do

    so then the derivative is

    0=\cos^3\theta - \sin\theta2\cos\theta\sin\theta

    then i can factor out a cosine and get
    0=\cos^2\theta - 2\sin^2\theta

    i know u have to solve for theta, but i kinda forgot how to solve trig equations
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  8. #8
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    You can factor out only if it's \neq 0

    The better way to do is to factorise, and solve...
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  9. #9
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    so...

    0=\cos\theta(\cos^2\theta - 2\sin^2\theta)

    how do u then solve it?
    im sure you can factor out the L^3/4 pi because it cannot be zero
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  10. #10
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    Here, yes, you can factor it out.

    Now, replace \cos^2(x) by 1-\sin^2(x) and then solve for it.

    ab=0 \text{ iff } a=0 \text{ or } b=0
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  11. #11
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    answer right??

    35.26 degrees??

    thank you very much for the help!
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